Question

我无法在我的代码中找到错误，但$ q.defer（）。reject（）似乎无法正常工作。 defer.resolve工作得很好，甚至可以进入finally段，但是defer.reject（虽然不会抛出任何错误但是）不会返回then函数，也不会返回finally函数我正在使用angularjs版本1.5.8

// calling service 
DashboardFactory.getDashboard($scope.dashboardData.dsDetails.dashName).then(
                    function (response) {
                        debugger;
                        console.log('success');
                    },
                    function () {
                        console.log('reject');
                    }
                ).finally(function() {
                    debugger;
                    console.log('finally');
             });

// service 
                var defer = $q.defer();
                $timeout(function() {
                    $http(httpParams)
                        .success(function(data, status, headers, config) {
                            defer.resolve(data);
                        })
                        .error(function(error, status, headers, config) {
                            console.log('error');
                            defer.reject('error');
                        })
                }, 0);
                return defer.promise;

Answer 1

$q是一项服务，允许您以异步方式运行函数，并且可以使用promises。

但$http服务方法get post put ...已经返回承诺，因此您不需要它，但如果您真的坚持使用它，您可以像这样使用它：

<!DOCTYPE html>
<html>

  <head>
    <script data-require="angularjs@1.5.8" data-semver="1.5.8" src="https://opensource.keycdn.com/angularjs/1.5.8/angular.min.js"></script>
    <link rel="stylesheet" href="style.css" />
    <script>
      var app = angular.module('yourApp', []);
      app.service('YourService', function($http, $q) {
        var url = "https://jsonplaceholder.typicode.com/users/";
        this.getUserWithQ = function(id) {
          var deferred = $q.defer();
          $http.get(url+id).then(
            function(data) {
              deferred.resolve(data);
            },
            function(error) {
              deferred.reject(error);
            }
          );
          return deferred.promise;
        };
        this.getUserWithoutQ = function(id) {
          return $http.get(url+id);
        }
      })
      app.controller('YourCtrl', function($scope, YourService) {
        YourService.getUserWithQ(1).then(
          function(res) {
            $scope.userWithQ = res.data;
          },
          function(error) {
            //unused
          }
        );
        YourService.getUserWithQ(11).then(
          function(res) {
            //unused
          },
          function(error) {
            $scope.userWithQerror = error;
          }
        );
        //Just for the record
        YourService.getUserWithoutQ(1).then(
          function(res) {
            $scope.userWithoutQ = res.data;
          },
          function(error) {
            //unused
          }
        );
        YourService.getUserWithoutQ(11).then(
          function(res) {
            //unused
          },
          function(error) {
            $scope.userWithoutQerror = error;
          }
        );
      });
    </script>
  </head>

  <body ng-app="yourApp">
    <div ng-controller="YourCtrl">
      <div> 
        <div>
          With $q success:
          {{userWithQ}}
        </div>
        <br>
        <div>
          With $q error:
          {{userWithQerror}}
        </div>
      </div>
      <br>
      <div> 
        <div>
          Without $q success:
          {{userWithoutQ}}
        </div>
        <br>
        <div>
          With $q error :
          {{userWithQerror}}
        </div>
      </div>
    </div>
  </body>
</html>

我已将此api用作具有10个用户的示例https://jsonplaceholder.typicode.com/users/，并且11将产生错误。

如您所见，您可以使用$q.defer()和$http返回值获得相同的结果。

以下是plunker

angularjs .reject没有返回到then函数

1 个答案: